Trigonometric Ratios of Specific Angles

Trigonometric Ratios of Specific Angles

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We can determine the trigonometric ratios for the following five angles based on our existing knowledge of pure geometry: \({0^ \circ},\,{30^ \circ },\,{45^\circ },\,{60^\circ }\,\rm{and}\,{90^\circ}\). Let us see how:

Trigonometric Ratios of \({0^\circ }\,\rm{and}\,{90^\circ}\)

Consider a \(\Delta ABC\) which is right-angled at \(B\), such that \(\angle A\) is very small:

T-ratios of 0 and 90 degrees

✍Note: As \(\angle A\) is close to \({0^ \circ, }\) \(\angle C\) is close to \({90^ \circ. }\)

Now,

  • \(\sin A = \frac{{BC}}{{AC}}\) will be very small (close to 0), because \(BC\) is very small.
  • \(\cos A = \frac{{AB}}{{AC}}\) will be almost equal to 1, because the base \(AB\) is almost equal to the hypotenuse \(AC\).

Visualize what would happen if \(\angle A\) was made smaller and smaller until it became \({0^ \circ }\).

In this scenario,

  • \(\sin A\) would become exactly 0.
  • \(\cos A\)  would become exactly 1.

Thus, we conclude that

\[\sin {0^\circ } = 0,\,\,\,\,\,\cos {0^\circ } = 1\]

If we analyze the sine and cosine of \(\angle C\) in the same situation, we can conclude that

\[\sin {90^ \circ } = 1,\,\,\,\,\,\cos {90^ \circ} = 0\]

We summarize these findings below:

\[\boxed { \begin{gathered}
  \sin 0^\circ  = 0,{\text{ }}\sin 90^\circ  = 1 \hfill \\
  \cos 0^\circ  = 1,{\text{ }}\cos 90^\circ  = 0 \hfill \\ 
\end{gathered} }\]


Trigonometric Ratios of \({30^ \circ }\,\,\rm{and}\,\,{60^\circ }\)

Consider a \(\Delta ABC\), which is right-angled at \(B\). Suppose that \(AC\) is twice that of \(BC\):

T-ratios of 30 and 60 degrees

Let \(D\) be the mid-point of \(AC\). We can show in plane geometry that \(AD = CD = BD\), and as a consequence, \(\angle A = {30^ \circ}\) and \(\angle C = {60^ \circ}\). You are urged to prove this assertion once again.

⚡Tip: Produce \(BD\) to a point \(E\) such that \(BD = DE\) and then join \(AE\). After that, prove \(\Delta EDA\) and \(\Delta BDC\) congruent.

Now, we have:

\[\begin{align}\sin A = \sin {30^\circ } &= \frac{{BC}}{{AC}} \hfill \\ &= \frac{{BC}}{{2BC}} \hfill \\ &= \frac{1}{2}\end{align}\]
_____________________
\[\begin{align}\cos C = \cos {60^ \circ } = \frac{{BC}}{{AC}} = \frac{1}{2}\end{align}\]
_____________________
\[\begin{align}\cos A = \cos{30^\circ } &= \frac{{AB}}{{AC}} \hfill \\ &= \sqrt {1 - {{\left( {\frac{{BC}}{{AC}}} \right)}^2}} \hfill \\ &= \frac{{\sqrt 3 }}{2}\end{align}\]
_____________________
\[\begin{align}\sin C = \sin {60^\circ } = \frac{{AB}}{{AC}} = \frac{{\sqrt 3 }}{2}\end{align}\]
_____________________

To summarize,

\[\boxed {\begin{gathered}
  \sin 30^\circ  = \frac{1}{2},{\text{ }}\sin 60^\circ  = \frac{{\sqrt 3 }}{2} \hfill \\
  \cos 30^\circ  = \frac{{\sqrt 3 }}{2},{\text{ }}\cos 60^\circ  = \frac{1}{2} \hfill \\ 
\end{gathered} }\]


Trigonometric Ratios of \({45^\circ }\)

Consider \(\Delta ABC\) which is right-angled at \(B\), with \(\angle A = {45^ \circ}.\) We note that \(\angle C\) will also be \({45^ \circ}\) and the triangle will be isosceles:

T-ratios of 45 degrees

Note that \(\begin{align}AB = BC = \frac{{AC}}{{\sqrt 2 }}\end{align}\) (how?), and thus

  • \(\begin{align}\sin {45^\circ } = \,\frac{{BC}}{{AC}} = \frac{1}{{\sqrt 2 }}\end{align}\)
  • \(\begin{align}\cos {45^ \circ } = \frac{{AB}}{{AC}} = \frac{1}{{\sqrt 2 }}\end{align}\)

To summarize,

\[\boxed {\sin {45^ \circ } = \frac{1}{\sqrt2},\,\,\cos {45^\circ } = \frac{1}{\sqrt2}}\]


Trigometric table for these angles:

Now that we have calculated the sine and cosine values for the five specific angles, let us summarize these in a table along with the values of the other trigonometric ratios for these angles.

 

\[\theta \]

\[{0^\circ }\]

\[{30^\circ }\]

\[{45^ \circ }\]

\[{60^\circ }\]

\[{90^ \circ }\]

\[\sin \theta \]

0

\[\frac{1}{2}\]

\[\frac{1}{{\sqrt 2 }}\]

\[\frac{{\sqrt 3 }}{2}\]

1

\[\cos \theta \]

1

\[\frac{{\sqrt 3 }}{2}\]

\[\frac{1}{{\sqrt 2 }}\]

\[\frac{1}{2}\]

0

\[\tan \theta \]

0

\[\frac{1}{{\sqrt 3 }}\]

1

\[\sqrt 3 \]

N.D.

\[\text {cosec}\, \theta \]

N.D.

2

\[\sqrt 2 \]

\[\frac{2}{{\sqrt 3 }}\]

1

\[\sec \theta \]

1

\[\frac{2}{{\sqrt 3 }}\]

\[\sqrt 2 \]

2

N.D.

\[\cot \theta \]

N.D.

\[\sqrt 3 \]

1

\[\frac{1}{{\sqrt 3 }}\]

0

✍Note: N.D. stands for not defined.

For example, \(\tan {90^\circ }\) is undefined because

\[\begin{align}\tan {90^ \circ } = \frac{{\sin {{90}^ \circ }}}{{\cos {{90}^ \circ }}} = \frac{1}{0}\end{align}\]

is a mathematically undefined value (as division by 0 is not allowed).

Note: You don’t need to memorize this table fully. Just remember the sine values, these facts:

\[\sin {0^ \circ} = \cos {90^ \circ},\,\,\,\,\sin {30^\circ } = \cos {60^ \circ}\]

\[\sin {90^\circ} = \cos {0^ \circ },\,\,\,\,\sin {45^\circ } = \cos {45^ \circ }\]

The other values can then be determined easily when required by using reciprocal relations for trigonometric ratios.


Solved Examples:

Example 1: Find the value of \(E = \cos {45^ \circ }\cos {30^\circ } + \sin {45^\circ }\sin {30^\circ }\)

Solution: we have:

\[\begin{align}&E = \frac{1}{{\sqrt 2 }} \times \frac{{\sqrt 3 }}{2} + \frac{1}{{\sqrt 2 }} \times \frac{1}{2}\\&\,\,\,\,\, = \boxed {\frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}}\end{align}\]


Example 2: Find the value of

\[\begin{align} E = \frac{{{{\cot }^2}\,{{30}^ \circ } + 4\,{{\cos }^2}\,{{45}^ \circ} + 3\,\rm{cosec}{^2}\,{{60}^ \circ }}}{{\sec {{60}^ \circ} + \rm{cosec}\,{{30}^ \circ } - {{\tan }^2}\,{{60}^ 0}}}\end{align}\]

Solution: We have

\[\begin{align}E &= \frac{{{{(\sqrt 3 )}^2} + 4{{\left( \frac{1}{\sqrt{2}} \right)}^{2}} + 3{{\left(\frac{2}{{\sqrt 3 }}\right)}^2}}}{{2 + 2 - 9{{(\sqrt 3 )}^2}}}\\\,\,\,\,\, &= \frac{{3 + 2 + 4}}{{2 + 2 - 3}}\\
\\&\Rightarrow\boxed{E = 9}\end{align}\]

yesChallenge 1: Evaluate the following:

\[\frac{{5{{\cos }^2}60^\circ  + 4{{\sec }^2}30^\circ  - {{\tan }^2}45^\circ }}{{{{\sin }^2}30^\circ  + {{\cos }^2}30^\circ }}\]

⚡Tip: Use trigonometric table for the values and then simplify.


Example 3: Suppose that \(0 < A + B < {90^\circ}\) and \(A > B\). If \(\tan \,(A + B) = \sqrt 3 \,\,\rm{and}\,\,\tan \left( {A - B} \right)\, = \frac{1}{\sqrt 3} \) , find the values of \(A\) and \(B\).

Solution: We have:

\[\tan \left( {A + B} \right) = \sqrt 3 \Rightarrow A + B = {60^ \circ}\]

\[\tan \left( {A - B} \right) = \frac{1}{{\sqrt 3 }} \Rightarrow A - B = {30^\circ }\]

Solving these two equations gives \(\boxed{A = 45^\circ }\) and \(\boxed{B = 15^\circ }\).

enlightenedThink: Why is the fact that \({0^ \circ } < A + B < {90^ \circ }\,\,\rm{and}\,\,A > B\) significant? Did we make any use of this fact?

yesChallenge 2: If \(\sin (A - B) = \frac{1}{2},\cos (A + B) = \frac{1}{2}\) and \({0^\circ } < A + B \leq 90^\circ ,A > B,\) find the values of \(A\) and \(B\).


Example 4: If \(\begin{align}\frac{{\sin {{60}^ \circ} + \cos {{30}^ \circ }}}{{1 + \cos {{60}^ \circ } + \sin {{30}^ 0 }}} = \sin \theta \end{align}\)

Find the value of \(\theta \) .

Solution: We have:

\[\begin{align}\frac{{\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 3 }}{2}}}{{1 + \frac{1}{2} + \frac{1}{2}}} &= \sin \theta \\ \Rightarrow \frac{{\sqrt 3 }}{2} &= \sin \theta \\ 
\\\Rightarrow &\boxed {\theta = {60^ \circ}}\end{align}\]


Example 5: Later on, we will study a lot of trigonometric identities (equations which are satisfied by any angle, in general). One of these identities is the following:

\[\tan 2\theta = \frac{{2tan\;\theta }}{{ 1- {{\tan }^2}\theta }}\]

Verify this identity using \(\theta = {30^\circ }\).

Solution: The L.H.S. is,

\[\tan 2\theta = \tan {60^ \circ } = \sqrt 3 \]

Substituting \(\theta = {30^ \circ}\) on the right side, we have

\[\begin{align}RHS = \frac{{2 \times \frac{1}{{\sqrt 3 }}}}{{1 - {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}} = \frac{{\frac{2}{{\sqrt 3 }}}}{{\frac{2}{3}}} = \sqrt 3\end{align}\]

The two sides are thus equal.
Hence Verified.


Example 6: Using \(A = {60^ \circ } \rm{and}\;B = {30^ \circ },\) verify the following identity:

\[\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B\]

Solution: We have:

\[\begin{align}&LHS = \sin {30^ \circ } = \frac{1}{2}\\&RHS = \frac{{\sqrt 3 }}{2} \times \frac{{\sqrt 3 }}{2} - \frac{1}{2} \times \frac{1}{2}\\&\;\;\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{4} - \frac{1}{4}\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\end{align}\]

The two sides are thus equal.
Hence Verified.


Example 7: Alpha is standing 20 m away from a building. There is a signboard somewhere on the building, whose angle of elevation from the alpha position is \({30^ \circ }\), while the angle of elevation of the top of the building is \({60^\circ}\). How much height difference is there between the signboard and the top of the building?

Solution: Observe the following figure carefully:

Height difference of two right-angled triangles

We have:

\[\frac{{{h_1}}}{{20}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \Rightarrow {h_1} = \frac{{20}}{{\sqrt 3 }}m\]

\[\frac{{{h_2}}}{{20}} = \tan {60^ \circ } = \sqrt 3 \Rightarrow {h_2} = 20\sqrt 3 m\]

Thus, the required height difference is,

\[\begin{align}&{h_2} - {h_1} = 20\sqrt 3 - \frac{{20}}{{\sqrt 3 }}\\&\,\,\,\,\,\,\,\,\,\,\;\,\,\,\,\,\,\,\, = \boxed{\frac{{40}}{{\sqrt 3 }} \approx 23.1m}\end{align}\]


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